In the previous section we completed the discussion on Division of lines. In this section, we will see Triangle division.
Consider the triangle ABC in fig.18.25(a) below. A red line is drawn parallel to the base AB. This red line intersect the sides AC and BC at P and Q.
At a first glance there is
nothing special about this fig(a). But some important details are hidden
in it. Let us bring them out:
Consider the triangle ABC in fig.18.25(a) below. A red line is drawn parallel to the base AB. This red line intersect the sides AC and BC at P and Q.
 |
| Fig.18.25 |
Draw two more parallel red
lines. One through the apex C, and another through the base AB. This
is shown in the fig(b). Now we have three parallel red lines. They cut
through two lines AC and BC. So the distances cut will be in the same
ratio. So we can write: AP:CP = BQ:CQ
This gives us a method to draw
a line parallel to the base of any triangle. Let us see an example:
Solved example 18.9
In the triangle ABC in the
fig.18.26(a) below, AC = 10 cm, BC = 15 cm and AP = 4 cm. Draw a line
parallel to AB through P
Solution:
 |
| Fig.18.26 |
• To draw a parallel to AB
through P, we need the perpendicular distance from P to AB. But it is
not given.
• What we have, is the distance
AP. Let us check whether there is any relation between this distance
and the total distance AC:
• We have: AP⁄AC = 4⁄10 = 2⁄5 ⇒ AP = 2⁄5 × AC.
• That means., AP is two fifths
of AC. In other words, if we divide AC into 5 equal parts, AP will
take up two such parts. It follows that, CP will take up the remaining
three parts.
• Thus we can say: P divides AC
in such a way that AP: CP = 2:3
• Now, we need a point Q on BC,
which will divide BC in such a way that BQ:CQ = 2:3. Let us find this
'Q':
• We have: BQ:CQ = 2:3. So if BC is divided into 5
equal parts, BQ should take up 2 such parts. In other words, BQ must
be two fifths of BC
• Thus we get: BQ = 2⁄5 × BC = 2⁄5 × 15 = 6 cm
• With B as centre, and 6 cm as
radius, draw an arc. This is shown in fig(b). This arc will cut BC at the required point Q
• Join P and Q. The Line PQ will
be parallel to AB
In the above problem, we
effectively used theorem 18.1 to draw the required parallel line. We found that
P divides AC in the ratio 2:3.
■ In fact we can use the method for any
ratio. Because, we have learned how to divide any line in any given
ratio.
■ So what if P is the exact
midpoint of AB?
• Then the calculations steps
are even more easier. The ratio is a simple 1:1. All we have to do is to draw a perpendicular
bisector of the other side BC. This bisector will give us the point
Q, and we can draw PQ
• From this we get some new
information. We can write it in the form of a theorem. We will write
it in steps.
Theorem 18.4
1. We have a triangle ABC with
base AB, and two sides AC and BC
2. We mark the midpoint of AC
as P, and the midpoint of BC as Q
3. Then we join PQ. This PQ
will be parallel to the base AB
We can write the converse
also:
1. We have a triangle ABC with
base AB, and two sides AC and BC
2. We mark the midpoint of AC
as P
3. Then we draw a line through
P, parallel to the base AB. This parallel line meets the other side
BC at Q
4. Q will be the midpoint of
BC
• Note that, in the triangle, any side can be taken as the 'base'. The 'two sides' will change accordingly.
■ The above steps can be written in just one line as:
In any triangle, the line drawn parallel to one side, passing through the midpoint of another side, meets the third side also at it's midpoint
Let us join the
midpoints of all the 3 sides of a triangle. This is shown in the fig.18.27(a) below:
• Note that, in the triangle, any side can be taken as the 'base'. The 'two sides' will change accordingly.
■ The above steps can be written in just one line as:
In any triangle, the line drawn parallel to one side, passing through the midpoint of another side, meets the third side also at it's midpoint
 |
| Fig.18.27 |
1. P is the midpoint of AC; Q is
the midpoint of BC; R is the midpoint of AB
2. Joining P, Q and R, we get an
inner triangle PQR
3. Also we have: PQ parallel to
AB (∵ P is midpoint of AC & Q is midpoint of BC, using theorem 18.4)
4.QR parallel to AC (∵ Q is
midpoint of BC & R is midpoint of AB)
5. PR parallel to BC (∵P is
midpoint of AC & R is midpoint of AB)
• Based on the above 5 points
on fig.18.27(a), we can derive some very interesting results:
6. We have AC parallel to QR ⇒ AP parallel to QR
7. Also we have AB parallel to PQ ⇒ AR parallel to PQ
• Let us take out this portion
ARQP for analysis. It is shown separately in fig.18.27(b)
8. We have two parallel lines PQ
and AR cut by a transversal PR. So ∠ARP = ∠RPQ
• They are alternate interior angles as shown in fig.18.27(c) below:
• The equal angles ∠ARP and ∠RPQ are shown in green colour
• They are alternate interior angles as shown in fig.18.27(c) below:
 |
| Fig.18.27 |
9. We have two parallel lines AP
and QR cut by a transversal PR. So ∠APR = ∠PRQ
• They are alternate interior angles as shown in fig.18.27(c) above.
• The equal angles ∠APR and ∠PRQ are shown in white colour
• They are alternate interior angles as shown in fig.18.27(c) above.
• The equal angles ∠APR and ∠PRQ are shown in white colour
10. Now consider the line PR. What
are the angles at it's end, from a point of view of ΔPQR?
Ans: The angles are ∠RPQ (green) and ∠PRQ (white)
11. What are the angles at it's
end, from a point of view of ΔAPR?
Ans: The angles are ∠APR (white) and ∠ARP (green)
12. But from (9), ∠APR (white) = ∠PRQ (white), and from (8), ∠ARP (green) = ∠RPQ (green)
• Now ask the question (11) again:
13. What are the angles at the
ends of line PR, from a point of view of ΔAPR?
Ans: The angles are ∠PRQ (white) and ∠RPQ (green)
14. The answers in (10) and (13) are the same. Line PR has green angle and white angle at it's ends in both the triangles. Thus. we have a line PR,
and the angles at it's ends, present in both ΔPQR and ΔAPR
15. It is a case of ASA
congruence (Angle, Side, Angle). The two triangles are equal. That is: ΔPQR = ΔAPR
16. Also, both the pairs of opposite sides are
parallel. So ARQP is a parallelogram
17. We proved (15) and (16), by taking out
the portion ARQP
18. In the same way, by taking out
the portion PRBQ, we can prove:
(i)ΔPQR = ΔQRB and (ii) PRBQ is a
parallelogram
19. In the same way, by taking out
the portion PRQC, we can prove:
(i) ΔPQR = ΔPQC and (ii) PRQC is a
parallelogram
20. From (15), (18.i) and (19.i), we can write:
ΔPQR = ΔAPR = ΔQRB = ΔPQC
■ That is., all the four inner
triangles are equal.
Another interesting result:
21. Consider the non-isosceles
trapezium ABQP
(i) AR = PQ. [since from (16), ARQP is a parallelogram, and AR and PQ are it's opposite sides]
(ii) Also, from (1), R is the midpoint of AB
(ii) Also, from (1), R is the midpoint of AB
• It follows that, PQ = AR = RB.
That means, PQ is half of AB
22. In a similar way, by
considering the non-isosceles trapezium BCPR, we get: PR is half of BC
23. In a similar way, by considering the
non-isosceles trapezium ACQR, we get: QR is half of AC
• We can write the above findings (21), (22) and (23) in the form of a theorem. In fact it is an extension of theorem 18.4
Theorem 18.5
1. We have a triangle ABC with
base AB, and two sides AC and BC
2. We mark the midpoint of AC
as P, and the midpoint of BC as Q
3. Then we join PQ. This PQ
will be parallel to the base AB
4. Also PQ will be half of AB
We can write the converse
also:
1. We have a triangle ABC with
base AB, and two sides AC and BC
2. We mark the midpoint of AC
as P
3. Then we draw a line through
P, parallel to the base AB. This parallel line meets the other side
BC at Q
4. Q will be the midpoint of
BC
5. Also PQ will be half of AB
• Note that, in the triangle, any side can be taken as the 'base'. The 'two sides' will change accordingly.
■ The above steps can be written in just one line as:
The length of the line joining the midpoints of two sides of a triangle is half the length of the third side
• Note that, in the triangle, any side can be taken as the 'base'. The 'two sides' will change accordingly.
■ The above steps can be written in just one line as:
The length of the line joining the midpoints of two sides of a triangle is half the length of the third side
Solved example 18.9
Fig.18.28 below shows a right angled triangle ABC. The midpoint of AC is marked as D. A perpendicular DE is dropped from D to the base AB. (i) Calculate the side CB of the larger triangle ABC. (ii) Calculate all the sides AE, AD and ED of the smaller triangle AED
 |
| Fig.18.28 |
Part (i): 1. ABC is a right angled triangle, as indicated by the 90o at the corner B. We can simply apply Pythagoras theorem to ⊿ABC:
CB2 = 102 - 82 ⇒ CB2 = 100 - 64 ⇒ CB2 = 36 ⇒ CB = √36 = 6 cm
Part(ii): 1. DE is parallel to BC. Because, both DE and BC are perpendicular to AB. Also, given that D is the midpoint of AC.
2. Theorem 18.4 states: In any triangle, the line drawn parallel to ones side, passing through the midpoint of another side, meets the third side also at it's midpoint.
3. So E is the midpoint of AB. Thus we get AE = 8⁄2 = 4 cm
4. Now we apply theorem 18.5: The length of a line joining the midpoints of two sides of a triangle is half the length of the third side.
5. So DE is half of BC. Thus we get DE = 6⁄2 = 3 cm
6. Given that D is the midpoint of AC. So DE = 10⁄2 = 5 cm
7. (3), (5) and (6) gives the answer to part (ii)
In the next section, we will see more details about Triangle Division.
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